The solution set of the equation 2sin2x -3cos x -1-0 is 2n-a-b-n - ab-0-1- where a-b are co-prime- Then the value of a-b is

2sin^2x +√(3)cos x +1=0 ⇒ 2(1 cos^2x)+√(3)cos x +1=0 ⇒ 2cos^2x √(3)cos x 3=0 ⇒(2cos x+√(3)) ( cos x √(3))=0 ⇒cos x= √(3)2 or cos x=√(3) nbsp; (not possible) ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiple of an Angle

The solution ...

Question

The solution set of the equation $2 {sin}^{2} x + \sqrt{3} cos x + 1 = 0$ is ${2 n π \pm \frac{a π}{b}, n \in Z; \frac{a}{b} \in [0, 1]}$ where $a, b$ are co-prime. Then the value of $a + b$ is

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Solution

The correct option is B $11$
$2 {sin}^{2} x + \sqrt{3} cos x + 1 = 0 \Rightarrow 2 (1 - {cos}^{2} x) + \sqrt{3} cos x + 1 = 0 \Rightarrow 2 {cos}^{2} x - \sqrt{3} cos x - 3 = 0 \Rightarrow (2 cos x + \sqrt{3}) (cos x - \sqrt{3}) = 0$
$\Rightarrow cos x = - \frac{\sqrt{3}}{2}$ or $cos x = \sqrt{3}$ (not possible)
$\Rightarrow cos x = cos \frac{5 π}{6}$
$\Rightarrow x \in {2 n π \pm \frac{5 π}{6}, n \in Z}$
$∴ a = 5$ and $b = 6$
$\Rightarrow a + b = 11$

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The solution set of the equation 2sin2x+√3cosx+1=0 is {2nπ±aπb,n∈Z; ab∈[0,1]} where a,b are co-prime. Then the value of a+b is

The correct option is B 112sin2x+√3cosx+1=0⇒2(1−cos2x)+√3cosx+1=0⇒2cos2x−√3cosx−3=0⇒(2cosx+√3)(cosx−√3)=0 ⇒cosx=−√32 or cosx=√3 (not possible) ⇒cosx=cos5π6 ⇒x∈{2nπ±5π6,n∈Z} ∴a=5 and b=6 ⇒a+b=11

The solution set of the equation $2 {sin}^{2} x + \sqrt{3} cos x + 1 = 0$ is ${2 n π \pm \frac{a π}{b}, n \in Z; \frac{a}{b} \in [0, 1]}$ where $a, b$ are co-prime. Then the value of $a + b$ is